Yesterday, on March 14th, nosotros celebrated Einstein's birthday equally good equally the Pi Day (3/14). You could convey bought Wolfram Mathematica at a discount, peculiarly because it was the rounded \(\pi\) day, 3.1416.
In Deutschland as well as countries influenced past times Germany, nosotros oft utilization the term "Ludolph number" for \(\pi\) – the "Archimedes constant", equally the English-speaking the world tends to enjoin – because 400 years ago, the manager of a fencing schoolhouse as well as mathematics professor Ludolph van Ceulen calculated twenty digits of \(\pi\) past times approximating it amongst the regular 1,073,741,284-gon after he decided that he wanted to convey a cool tombstone (some digits were added later). Yes, the release of the sides is \(2^{30}\) as well as so y'all may larn the lawsuit past times "halving as well as halving thirty times" – this method was already known to Archimedes 1700 years before – as well as the divergence of the circumference of this "billiongon" from the circle goes similar "one over 1 billion squared". About twenty digits, indeed.
The release \(\pi\) is the most pop irrational release amid the laymen. They associate diverse supernatural properties amongst it – equally y'all could run into inwards the thriller named Pi. Much of this hype is irrational (much similar \(\pi\), pun intended) because all other irrational or transcendental numbers – as well as almost all existent numbers are irrational as well as transcendental – portion those properties (similar to irrationality) that the laymen are capable of describing past times words (e.g. their digits are difficult to remember). Some of the hype is justified because \(\pi\) genuinely appears at dainty places of mathematics.
Feynman's #1 favorite identity of mathematics was Euler's identity\[
\exp(i \pi ) + 1 = 0.
\] Because the exponentiation of purely imaginary numbers produces rotations of the complex plane, this formula is linked amongst the commons circumference of the circle, \(2\pi r\).
The release \(\pi\) appears inwards lots of mathematical identities that are non "obviously" related to the circumference of circles or surfaces or volumes of spheres. For example, the integral of the Gaussian function\[
\int_{-\infty}^{+\infty} dx\, \exp(-x^2) = \sqrt{\pi}.
\] Why is this true? Well, the circle isn't manifestly introduce inwards the integral but the circle does appear inwards a clever calculation of the integral. Square the integral as well as y'all volition get\[
\int_{-\infty}^{+\infty} dx\,dy \exp(-x^2-y^2) =\dots
\] which is rotationally symmetric. You may switch to the polar coordinates. The integral over \(\phi\) produces \(2\pi\) (that's how nosotros cleverly got the "circle" dorsum although it wasn't inwards the master integral) piece the remaining portion (in which nosotros don't forget the portion of \(r\) from the Jacobian compulsory for the polar coordinates) is the radial integral\[
\int_0^\infty dr\, r\,\exp(-r^2) = -\frac{1}{2} \left. \exp(-r^2)\right|_0^\infty = \frac 12
\] The full production is \(2\pi \times 1/2 = \pi\). The master integral must live the foursquare root of that as well as it was positive, thence \(\sqrt{\pi}\).
In an article almost a sure hype almost \(\pi\) as well as the hydrogen atom, I discussed the fact that \((-1/2)!=\sqrt{\pi}\), too. It is genuinely equal to the same integral nosotros merely calculated – but at that topographic point are many other ways to await at it.
And Wolfram Mathematica at a discount, peculiarly because it was the rounded \(\pi\) day, 3.1416.
In Deutschland as well as countries influenced past times Germany, nosotros oft utilization the term "Ludolph number" for \(\pi\) – the "Archimedes constant", equally the English-speaking the world tends to enjoin – because 400 years ago, the manager of a fencing schoolhouse as well as mathematics professor Ludolph van Ceulen calculated twenty digits of \(\pi\) past times approximating it amongst the regular 1,073,741,284-gon after he decided that he wanted to convey a cool tombstone (some digits were added later). Yes, the release of the sides is \(2^{30}\) as well as so y'all may larn the lawsuit past times "halving as well as halving thirty times" – this method was already known to Archimedes 1700 years before – as well as the divergence of the circumference of this "billiongon" from the circle goes similar "one over 1 billion squared". About twenty digits, indeed.
The release \(\pi\) is the most pop irrational release amid the laymen. They associate diverse supernatural properties amongst it – equally y'all could run into inwards the thriller named Pi. Much of this hype is irrational (much similar \(\pi\), pun intended) because all other irrational or transcendental numbers – as well as almost all existent numbers are irrational as well as transcendental – portion those properties (similar to irrationality) that the laymen are capable of describing past times words (e.g. their digits are difficult to remember). Some of the hype is justified because \(\pi\) genuinely appears at dainty places of mathematics.
Feynman's #1 favorite identity of mathematics was Euler's identity\[
\exp(i \pi ) + 1 = 0.
\] Because the exponentiation of purely imaginary numbers produces rotations of the complex plane, this formula is linked amongst the commons circumference of the circle, \(2\pi r\).
The release \(\pi\) appears inwards lots of mathematical identities that are non "obviously" related to the circumference of circles or surfaces or volumes of spheres. For example, the integral of the Gaussian function\[
\int_{-\infty}^{+\infty} dx\, \exp(-x^2) = \sqrt{\pi}.
\] Why is this true? Well, the circle isn't manifestly introduce inwards the integral but the circle does appear inwards a clever calculation of the integral. Square the integral as well as y'all volition get\[
\int_{-\infty}^{+\infty} dx\,dy \exp(-x^2-y^2) =\dots
\] which is rotationally symmetric. You may switch to the polar coordinates. The integral over \(\phi\) produces \(2\pi\) (that's how nosotros cleverly got the "circle" dorsum although it wasn't inwards the master integral) piece the remaining portion (in which nosotros don't forget the portion of \(r\) from the Jacobian compulsory for the polar coordinates) is the radial integral\[
\int_0^\infty dr\, r\,\exp(-r^2) = -\frac{1}{2} \left. \exp(-r^2)\right|_0^\infty = \frac 12
\] The full production is \(2\pi \times 1/2 = \pi\). The master integral must live the foursquare root of that as well as it was positive, thence \(\sqrt{\pi}\).
In an article almost a sure this article gave y'all a proof of the Machin formula\[
\pi = 16\arctan \frac{1}{5} - 4 \arctan \frac{1}{239}.
\] which allows y'all to calculate \(\pi\) past times Taylor serial much to a greater extent than rapidly than the "simpler" \(\pi = 4\arctan 1\). I could cash inwards one's chips along amongst interesting identities involving \(\pi\) for an hr or a few hours. You would listen almost diverse sums that may live interpreted equally Taylor serial for trigonometric as well as related functions; or the zeta business office of fifty-fifty positive integers which contains \(\pi^{2s}\) because 1 may calculate them using the norms of some functions as well as their Fourier series, as well as and so on.
But allow me boundary to something to a greater extent than interesting or to a greater extent than controversial, namely "seemingly divergent" products as well as sums.
In lots of previous weblog posts, I discussed the essence of positive integers,\[
1+2+3+4+5+\dots = -\frac{1}{12}.
\] This lawsuit plant assuming the zeta-function regularization of the superficially divergent sum; but the same lawsuit may genuinely live obtained past times regularization procedures of many dissimilar types. This identity is "really true" inwards some deep feel – as well as it's useful e.g. for the simplest evaluation of the critical spacetime dimension inwards string theory, \(D=26\) for bosonic string theory or \(D=10\) for the superstring.
The essence of positive integers ends upward beingness equal to \(\zeta(-1)\) as well as this Riemann zeta business office may live extrapolated to all complex values of the declaration – it is finite everywhere except for the pole at \(s=1\). Similarly, 1 may evaluate\[
1^s+2^s+3^s+\dots = \zeta(-s).
\] This essence of powers of positive integers is aught for fifty-fifty positive values of \(s\); it is nonzero for strange positive values of \(s\). For example, the essence of 3rd powers is \(\zeta(-3)=+1/120\). Now, tin y'all also evaluate the finite value of the production of all positive integers?\[
1\times ii \times 3\times 4\times \dots = ?
\] Yes, y'all can. First, this production may live converted to the exponential\[
\dots = \exp (\ln 1 + \ln ii + \ln 3+\dots) = ?
\] as well as the remaining chore is to observe the regularized "sum of logarithms of integers". The reply may live obtained past times looking at the zeta function, too. Recall that\[
\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}
\] as well as the logarithmic prefactor may live easily produced if nosotros differentiate amongst honour to \(s\):\[
\zeta'(s) = -\sum_{n=1}^{\infty} \frac{\ln n}{n^s}
\] We may eliminate the \(1/n^s\) portion past times substituting \(s=0\). So to observe the "sum of the logs of integers", nosotros postulate to know \(-\zeta'(0)\). By the \(s\leftrightarrow 1-s\) functional equation, the demeanour of \(\zeta\) around \(s=0\) is linked to the demeanour around \(s=1\) where the master essence formula for \(\zeta\) is convergent. One may calculate \(\zeta'(0)\) as well as \(\zeta'(1)\) analytically (the latter has a divergent term) but it's straightforward as well as the "finite" calculations are non mysterious. So why don't nosotros merely inquire Mathematica?
The symbolic as well as numerical answers agree:
a = 0.000000001; -(Zeta[a] + 0.5)/aSo yes, \(-\zeta'(0)\) nosotros wanted to know is \(\ln \sqrt{2\pi}\) (note that the zeta business office "knows" that the derivative is equal to the logarithm of something simpler) which implies that the production of all positive integers is\[
-D[Zeta[s], s] /. {s -> 0}
0.918939
\(\log(2\pi)/2\)
1\times 2\times 3\times \dots = \sqrt{2\pi}.
\] You could convey gotten this lawsuit past times an fifty-fifty faster but sloppier, ultra-heuristic method. This production is clearly the factorial of a number, \(\infty!\), as well as for such large numbers, 1 may utilization the Stirling's approximation\[
n! \approx \sqrt{2\pi n} \zav { \frac{n}{e} }^n
\] Now, the ability as well as the polynomial prefactors are divergent for \(n\to \infty\) as well as difficult to interpret. But these factors await "too simple" as well as multiplicatively, the solely reasonable matter that such expressions may live equal to is one. On the other hand, the \(\sqrt{2\pi}\) prefactor is at that topographic point as well as can't live ignored, as well as so it's the lawsuit for \(\infty! = \sqrt{2\pi}\).
Euler has succeeded inwards rewriting the zeta business office equally a production over primes. That's why 1 may observe the regularized value of the production over primes only, non all integers. You may await e.g. at the 2003 newspaper past times Muñoz-Garcia as well as Pérez Marco; they wrote a followup 5 years later.
Their funny lawsuit for the production of primes – obtained past times a similar conversion of the production to values of the \(\zeta\) function, its powers, derivatives, as well as production as well as ratios of such objects – reads\[
2\times iii \times 5 \times vii \times \dots = 4\pi^2.
\] The production of all primes is "four pi squared". It is the 4th ability of the production of all positive integers, a fact that may in all likelihood live proven inwards seemingly independent ways. With the usage of the Euler production as well as the previous lawsuit (no "new" regularization is needed), 1 may also calculate products of \((p^s-1)\) for whatever value of \(s\) as well as the lawsuit is\[
\prod_p (p^s - 1) = \frac{(2\pi)^{2s}}{\zeta(s)}.
\] For example, the production of \((p-1)\) over primes happens to live aught because \(\zeta(1)\) has the pole. Similarly,\[
\prod_p (p^2 - 1) = 48\pi^2.
\] In fact, for all fifty-fifty positive values of \(s\), the production of \((p^s-1)\) over primes is, merely similar \(\zeta(s)\) itself, a rational multiple of \(\pi^s\).
I believe that much of mathematics would cash inwards one's chips to a greater extent than natural – as well as some dainty insights (perhaps fifty-fifty unknown ones) would cash inwards one's chips to a greater extent than easily accessible – if the zeta business office (and/or similar, full general as well as skillful enough) regularization were implicitly used everywhere. For example, I believe that it makes feel to enjoin that\[
\dots (t-2)(t-1)t(t+1)(t+2)\dots = C \sin \pi t
\] where the constant \(C\) may live determined past times some consistency checks as well as volition in all likelihood include the portion of \(\zeta(-1)=-1/12\). The sign may live undetermined. Or is some exponential portion missing inwards my result? Similarly, the product\[
t(t+1)(t+2)\dots = C' \Gamma(-t).
\] Just don't live afraid to write many of these products – which are singular without regularization – because the finite value associated amongst them past times all "good plenty regularization schemes" is uniquely determined. I practise think that such regularized values could live useful for accelerating noesis as well as proofs – inwards release theory, the interrogation into the Riemann Hypothesis, physics, as well as beyond.
Bonus: Ludolph's calculation
Many people click at Ludolph's tombstone inwards a higher house as well as so allow me enjoin a few words almost the calculation. Draw a regular \(n\)-gon within a unit of measurement circle where \(n=2^{30}\). You may run into that the circumference of this polygon is a chip shorter than \(2\pi\). The circumference is \(2^{30}\) times the side, as well as the side is \(2\times \sin(\pi/2^{30})\). So the circumference is \(2^{31} \sin (\pi/2^{30})\).
Influenza A virus subtype H5N1 funny matter is that \(\sin(\pi/2^{30})\) may live calculated past times halving the angles many times. You iterate the formula \(\sin(2y)=1-2\sin^2 y\) i.e.\[
\sin(x/2) = \sqrt{\frac{1-\sin x}{2}}
\] as well as because y'all know how to calculate the foursquare roots numerically as well as because y'all know that \(\sin(\pi/4)=\sqrt{1/2}\), y'all may larn an accurate lawsuit after some numerical calculation that is doable without a computer. Numerically, \(2^{31} \sin (\pi/2^{30})\) is almost \(2\pi - 8.96455\times 10^{-18}\).
You may also consider the \(n\)-gon for \(n=2^{30}\) that contains the circle inside. This 1 has a longer circumference than \(2\pi\) as well as this circumference is also calculable. It's \(\cos(\pi/2^{30})\) times longer than the previous 1 – as well as this cosine is also numerically calculable because \(\cos^2 x = 1-\sin^2 x \) – i.e. \(2^{31}\tan(\pi/2^{30})\). This ends upward beingness equal to \(2\pi+1.7929105\times 10^{-17}\) or so.
If y'all were clever plenty (and Ludolph himself wasn't, I think), y'all could figure out that a much amend approximation for \(2\pi\) is the weighted average – taking the sine (underestimate) amongst the weight \(2/3\) as well as the tangent (overestimate) amongst the weight \(1/3\). If y'all practise so, the weighted average volition live \(2\pi+2.3\times 10^{-35}\) or as well as so – 35 digits of \(2\pi\) volition live right (exactly plenty for Ludolph's modernized tombstone).
There for sure be to a greater extent than schoolkid-friendly ways to demo that the optimal weights are \(2/3\) as well as \(1/3\) but the simplest proof for adults, I think, is to notice that \(\sin z= z-z^3/6+\dots\) as well as \(\tan z = z+z^3/3+\dots\) piece nosotros would genuinely similar to include "just" the term \(z\) to larn \(\pi\). The fault \(-z^3/6\) is twice equally pocket-sized equally the fault of \(+z^3/3\), as well as so if the sine gets twice equally high a weight than the tangent, the errors from the \(z^3\) price (around \(10^{-18}\) inwards our calculation of \(\pi\)) cancel as well as solely the \(z^5\) price survive, giving the fault \(10^{-35}\) to \(\pi\).